Solution:

The two possible positions of $D$ are symmetric with respect to the line $(AB)$, so it suffices to consider the case in which the triangle $ABD$ is oriented counterclockwise (see figure). Let the common center of the two circles be denoted by $M$. Since the tangent radius $MD$ is perpendicular to the tangent $BC$, the triangle $MBD$ is right-angled, so that by the Pythagorean theorem

$|BD|^{2}=|MB|^{2}-|MD|^{2}=r_{1}^{2}-r_{2}^{2}=169-64=105$ follows.

By Thales' theorem, $\text{ACB=90}$. Since, because of $\text{ACB= MDB=90}$ and the common angle $\text{DBM= CBA}$, the triangles $ABC$ and $MBD$ are similar, and since $|MB|=r_{1}=|MA|$ holds, it also follows that $|DC|=|BD|=\sqrt{105}$ as well as $|CA|=2 \cdot|DM|=16$. Thus the lengths of the legs in the right triangle $ADC$ are known and it follows that $|AD|=\sqrt{105+16^{2}}=\sqrt{361}=19$. The side $AD$ therefore has length 19.